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Wiener-Khintchine theorem and negative frequencies
Hi,
What is the meaning of a negative frequency? (in the context of the spectrum of a stochastic process) I have seen time to time that we extend integration to negative frequencies to benefit from Fourier transform calculus, and that it was virtually the same from a physical point of view because the contributions of negative (unphysical) frequencies were negligible. However if you compute the power spectrum S(omega) of the random telegraph signal, you get for S a Lorentzian centered about 0, i.e., fluctuations of frequencies -omega_0 are as important as +omega_0, so no question here to say that unphysical negative frequencies are negligibles. For this special case of telegraph signal, how do you understand its spectrum? Why is it centered about 0? What is the meaning of this zero frequency (no oscillations; I'd think the main frequency, i.e., the peak of the spectrum, should be centered on the inverse rate of transition)? Why are there negative frequencies on an equal footing with positive ones? Thanks. |
Wiener-Khintchine theorem and negative frequencies
Student wrote in message ...
Hi, What is the meaning of a negative frequency? (in the context of the spectrum of a stochastic process) When looking at the definition of the Fourier transform, you see that your signal is decomposed in a superposition of functions of the form exp(-j w t), where w goes from -infinity to + infinity. When you look at a definite frequency, you should in fact take into account that frequency and its negative: exp(-j w t) and exp (+ j w t). The sum of both gives you a real function, namely Cos(w t). The difference of both gives you another real function, namely Sin(w t). So if you combine exp(-j w t) and exp(+j w t) with equal weight, but opposite complex phases, you get a real function Cos(w t + phi). The only way to get REAL combinations is that the coefficient of the "negative frequency" and the one of the "positive frequency" are of equal magnitude and opposite complex phase, meaning they are conjugate. So the Fourier transform of a real signal is always a function that has the property: H(w) = H(-w)* (so what you write that the negative frequencies in real signals are neglegible is not true: they are just as present as the positive ones). Of course there is no extra information in the negative frequency part of the Fourier transform: if you know H(w) for w0, then you know it for w0, exactly by the relation given above. Concerning spectral power density and so on, it's just a matter of convention. Given the fact that |H(w)| = |H(-w)|, you can chose to just integrate over the positive frequencies and put a factor of 2 in front, or just integrate from -INfinity to +Infinity. cheers, Patrick. |
Wiener-Khintchine theorem and negative frequencies
In article , Student wrote: For this special case of telegraph signal, how do you understand its spectrum? Why is it centered about 0? What is the meaning of this zero frequency (no oscillations; I'd think the main frequency, i.e., the peak of the spectrum, should be centered on the inverse rate of transition)? Why are there negative frequencies on an equal footing with positive ones? Negative frequencies appear because you have a real-valued signal and you are taking a complex Fourier transform. If you did sine and cosine transforms you would find no conceptual problems, I presume. In general you expect the power of the positive and negative frequencies to be the same, so the spectrum will be symmetric about zero. I still expect you to find peaks away from zero, associated to physical frequencies or inverses of characteristic times. Regards, Miguel Carrion |
Wiener-Khintchine theorem and negative frequencies
The zero frequency is the average dc value of the signal. The negative
frequencies are a result of how the Fourier transform breaks down the signal. "Student" wrote in message ... What is the meaning of a negative frequency? (in the context of the spectrum of a stochastic process) |
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